Question 19988
Seems like you have missed the point of this problem.

{{{17i/(4-i) = (17i)(4+i)/(4-i)(4+i)}}} = {{{17(4i+i^2)/(16-i^2))}}}
Recall that: {{{i^2 = -1}}}, so:
{{{17(4i+i^2)/(16-i^2) = 17(4i-1)/(16+1)}}} = {{{17(4i-1)/17}}} = {{{(4i-1)}}} = {{{-1-4i}}} 
Compare this: -1-4i with A+Bi 
A = -1, the real part of the complex number.
Bi = -4i, the imaginary part of the complex number.